Assume that an LED of rating 3V and 20mA is connected in series with a resistance and a supply voltage of 9V is applied across it. For the required ratings of LED, I chosen a resistance value of 300 Ohms (Theoretical Value). Now, when I increased the supply voltage upto 12V by taking current 10mA~70mA, What will be the voltage across that LED and also the current through it? Actually the 3V LED will burnout but what will be the theoretical current and voltage across it when supply voltage is 12V?

  • 1
    \$\begingroup\$ sufficient current to burn it out... \$\endgroup\$
    – Solar Mike
    Aug 26, 2018 at 19:02
  • \$\begingroup\$ Is my calculation correct: Iled= (Vsource - Vled)/R \$\endgroup\$
    – Beginner
    Aug 26, 2018 at 19:08
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    \$\begingroup\$ It depends on the spec of the LED. Without that, it's unknown. Find a datasheet for your LED. \$\endgroup\$
    – adamaero
    Aug 26, 2018 at 19:09
  • 2
    \$\begingroup\$ Ask the question you want to ask, don't change the goal posts... \$\endgroup\$
    – Solar Mike
    Aug 26, 2018 at 19:10
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    \$\begingroup\$ This question cannot be answered because it is unclear when the LED is damaged. Some LEDs rated for 20 mA can handle 50 mA as long as you keep them cool enough. What would be the point of operating a LED rated for 20 mA at 30 mA? LED damage is not always as easy as that it "burns out", LEDs have several failure modes. \$\endgroup\$ Aug 26, 2018 at 19:17

1 Answer 1


enter image description here

Figure 1. A white LED I-V curve showing a nominal value and possible spread in values from device to device. Source: LED binning.

The forward voltage, VF, drop across an LED remains fairly constant over a range of current values. In your case we can assume that it is 3 V in the region of interest. Let's check ...


simulate this circuit – Schematic created using CircuitLab

Figure 2. The test circuit.

With 300 Ω in series on a 9 V supply there will be 6 V across the resistor. \$ I = \frac {V}{R} = \frac {6}{300} = 20 \ \text {mA} \$.

At 12 V supply there will be 9 V across the resistor so current will be up by 50% to 30 mA.

Looking at Figure 1 we can see that the voltage drop across the LED will not change significantly between 20 and 30 mA.


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